Time Limit: 10000MS | Memory Limit: 64000K | |
Total Submissions: 2553 | Accepted: 1232 | |
Case Time Limit: 2000MS | Special Judge |
Description
Two companies, Macrosoft and Microhard are in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove that the program in question is disgusting. However, Ivan has a complicated problem. This new program has s subcomponents, and finding bugs of all types in each subcomponent would take too long before the target could be reached. So Ivan and Microhard agreed to use a simpler criteria --- Ivan should find at least one bug in each subsystem and at least one bug of each category.
Macrosoft knows about these plans and it wants to estimate the time that is required for Ivan to call its program disgusting. It's important because the company releases a new version soon, so it can correct its plans and release it quicker. Nobody would be interested in Ivan's opinion about the reliability of the obsolete version.
A bug found in the program can be of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bug cannot belong to two different categories or happen simultaneously in two different subsystems. The number of bugs in the program is almost infinite, so the probability of finding a new bug of some category in some subsystem does not reduce after finding any number of bugs of that category in that subsystem.
Find an average time (in days of Ivan's work) required to name the program disgusting.
Input
Output
Sample Input
1 2
Sample Output
3.0000
Source
题意:一共有n类bug,s段 subcomponents,某人每天能够发现一个 subcomponent中的一个bug,求发现所有类bug并且每个 subcomponents中都有bug的期望。
思路:dp[i][j]表示找到i种bug,存在于j个subcomponents中的期望,则会发现,每当一天发现一个bug时,会有四种情况,将这些情况组合组合一下推出个公式带进去就好了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
using namespace std;
const int mod=1e9+7;
const int N=1005;
double dp[N][N];
int main(){
int n,s;
scanf("%d%d",&n,&s);
dp[n][s]=0;
for(int i=n;i>=0;i--){
for(int j=s;j>=0;j--){
if(i==n&&j==s)continue;
dp[i][j]=(i*(s-j)*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(n-i)*(s-j)*dp[i+1][j+1]+n*s)/(n*s-i*j);
}
}
printf("%.4f\n",dp[0][0]);
return 0;
}